Calculate the Gibbs free energy for the given reaction, with the standard electrode potential equal to 1.1 V.

$Zn(s) + Cu^{2+}(aq) →Zn^{2+}(aq) + Cu (s)$

$-212.27\, kJ/mol$

The correct answer is Option (3) → $-212.27\, kJ/mol$

For the reaction:

$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

Given:
E° = 1.1 V
Number of electrons transferred, n = 2
Faraday constant, F = 96,500 C/mol

Formula:

$\Delta G^\circ = -nFE^\circ$

Calculation:

$\Delta G^\circ = - (2)(96500)(1.1)$

$\Delta G^\circ = -212300 \, \text{J/mol} = -212.3 \, \text{kJ/mol}$

✅ Correct answer: −212.27 kJ/mol